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\begin{document}

 \author[Christoph Koutschan]{Christoph Koutschan\,$^1$}
 \address{Christoph Koutschan, Mathematics Department, Tulane University, New Orleans, LA, USA.}
% \address{Christoph Koutschan, Research Institute for Symbolic Computation, J. Kepler University Linz, Austria}
 \email{ckoutsch@risc.uni-linz.ac.at}
 \thanks{$^1$ Supported by NFS-DMS 0070567 as a postdoctoral fellow and by the Austrian Science Foundation (FWF) grant Y464-N18.}

 \author[Manuel Kauers]{Manuel Kauers\,$^2$}
 \address{Manuel Kauers, Research Institute for Symbolic Computation, J. Kepler University Linz, Austria}
 \email{mkauers@risc.uni-linz.ac.at}
 \thanks{$^2$ Corresponding author. Supported by the Austrian Science Foundation (FWF) grant Y464-N18.}
%\thanks{$^\ast$ Supported in part by the Austrian Science Foundation (FWF) grants P19462-N18 and P20162-N18.}

 \author[Doron Zeilberger]{Doron Zeilberger\,$^3$}
 \address{Doron Zeilberger, Mathematics Department, Rutgers University (New Brunswick), Piscataway, NJ, USA.}
 \email{zeilberg@math.rutgers.edu}
 \thanks{$^3$ Supported in part by the United States of America National Science Foundation.}

 \title{A~Proof of George~Andrews' and Dave~Robbins' $q$-TSPP~Conjecture}

\begin{abstract}
The conjecture that the orbit-counting generating function for totally
symmetric plane partitions can be written as an explicit
product-formula, has been stated independently by George Andrews and
Dave Robbins around 1983.  We present a proof of this long-standing
conjecture.
\end{abstract}

 \maketitle

\section{Proemium}

In the historical conference {\it Combinatoire \'{E}numerative} that took place
at the end of May 1985, in Montreal,
Richard Stanley raised some intriguing problems about the
enumeration of plane partitions (see below), which he later expanded into a
fascinating article~\cite{Sta1}. Most of these problems concerned the
enumeration of ``symmetry classes'' of plane partitions that were
discussed in more detail in another article of Stanley~\cite{Sta2}.
All of the conjectures in the latter article have since been proved
(see Bressoud's modern classic~\cite{B}), except one, which until now 
resisted the efforts of the greatest minds in enumerative
combinatorics.  It concerns the proof of an explicit formula for the
$q$-enumeration of totally symmetric plane partitions, conjectured,
ca. 1983, independently by George Andrews
and Dave Robbins~(\cite{Sta2}, \cite{Sta1} conj.~7, \cite{B}
conj.~13, and already alluded to in \cite{Andrews80}).  In the 
present article we finally turn this conjecture into a theorem.

A plane partition $\pi$ is an array $\pi=(\pi_{i,j})_{1\leq i,j}$, of
positive integers $\pi_{i,j}$ with finite sum $|\pi|=\sum \pi_{i,j}$,
which is weakly decreasing in rows and columns so that
$\pi_{i,j}\geq\pi_{i+1,j}$ and $\pi_{i,j}\geq\pi_{i,j+1}$. A plane
partition~$\pi$ is identified with its 3D Ferrers diagram which is
obtained by stacking $\pi_{i,j}$ unit cubes on top of the
location~$(i,j)$. This gives a left-, back-, and bottom-justified
structure in which we can refer to the locations $(i,j,k)$ of the
individual unit cubes. If the diagram is invariant under the action of
the symmetric group~$S_3$ then $\pi$ is called a totally symmetric
plane partition (TSPP). In other words, $\pi$ is called totally
symmetric if whenever a location $(i,j,k)$ in the diagram is occupied
then all its up to~$5$ permutations
$\{(i,k,j),(j,i,k),(j,k,i),(k,i,j),(k,j,i)\}$ are occupied as well.
Such a set of cubes, i.e., all cubes to which a certain cube can be
moved via $S_3$ is called an orbit; the set of all orbits of~$\pi$
forms a partition of its diagram (see Figure~\ref{tsppfigure}).

\begin{figure}\label{tsppfigure}
\includegraphics[width=75mm]{PP3D}
\caption{A totally symmetric plane partition with largest
part~$7$. The yellow cubes form an orbit of length~$6$ (corresponding
to all permutations of $(2,4,6)$), the red cubes an orbit of
length~$3$ (all permutations of $(1,1,7)$), and the green cube on
location $(4,4,4)$ represents an orbit of length~$1$.}
\end{figure} 

In 1995, Stembridge~\cite{Ste} proved Ian Macdonald's conjecture
that the number of totally symmetric plane partitions with largest part at most~$n$, i.e.,
those whose 3D Ferrers diagram is contained in the cube $[0,n]^3$,
is given by the elegant product-formula
\[
  \prod_{ 1 \leq i \leq j \leq k \leq n} \frac{i+j+k-1}{i+j+k-2}.
\]
Ten years after Stembridge's completely human-generated proof,
Andrews, Paule and Schneider~\cite{APS} came up with a
computer-assisted proof based on an ingenious matrix decomposition,
but since no $q$-analog of their decomposition was found, their
proof could not be extended to a proof for the $q$-case.  A third
proof of Stembridge's theorem~\cite{Koutschan10a}, completely
computer-generated, was recently found in the context of our
investigations of the $q$-TSPP conjecture. We have now succeeded in
completing all the required computations for an analogous proof of the
$q$-TSPP conjecture, and can therefore announce:

\begin{thm}\label{qTSPPthm}
Let $\pi/S_3$ denote the set of orbits of a plane partition~$\pi$
under the action of the symmetric group~$S_3$.  Then the
orbit-counting generating function (\cite[p.~200]{B},
\cite[p.~106]{Sta2}) is given by
\[
  \sum_{\pi\in T(n)} q^{|\pi/{S_3}|}=\prod_{1 \leq i \leq j \leq k \leq n}\frac{1-q^{i+j+k-1}}{1-q^{i+j+k-2}}
\] 
where $T(n)$ denotes the set of totally symmetric plane partitions
with largest part at most~$n$.
\end{thm}

\begin{proof}[Proof sketch] 
 Our proof is based on a result by Okada~\cite{O} who has shown that
 the theorem is implied by a certain---conjectured---determinant
 evaluation.  These preliminaries are stated accurately in Section~\ref{secReduction} below,
 followed by a description of the holonomic ansatz~\cite{Z3} that we
 are going to pursue.  This approach relies on a kind of oracle that
 tells us a description of a certificate function~$c_{n,j}$;
 the odyssey how this function has been ``guessed'' is described in Section~\ref{secGuess}. 
 Once $c_{n,j}$ is known, the determinant evaluation reduces further to
 proving the three identities \eqref{eq:Normalization},
 \eqref{eq:Soichi}, and \eqref{eq:Okada} stated below.  
 In Section~\ref{secProof} we come full circle by proving these identities.
 Additional technical details of the proof and our computations 
 as well as the explicit certificates are provided electronically 
 on our website (http://www.risc.uni-linz.ac.at/people/ckoutsch/qtspp/).
\end{proof}

Our proof is noteworthy not only for its obvious significance in enumerative combinatorics, as
it settles a long-standing conjecture, attempted by many people. It is noteworthy also for
computational reasons, as the computations we performed went far beyond what has
been thought to be possible with currently known algebraic algorithms,
software packages, and computer hardware.

\section{The Telemachiad}\label{secReduction}

In order to prove the $q$-TSPP conjecture, we exploit an elegant reduction
by Okada~\cite{O} to the problem of evaluating a certain determinant.
This determinant is also listed as Conjecture~46 in Krattenthaler's
essay~\cite{K} on the art of determinant evaluation.

Let, as usual, $\delta_{i,j}$ be the Kronecker delta function
and let, also as usual,
\[
  \qbinom n k = \frac{(1-q^n)(1-q^{n-1}) \cdots (1-q^{n-k+1})}{(1-q^k)(1-q^{k-1}) \cdots (1-q)}
\]
denote the $q$-binomial coefficient.
Define the discrete function $a_{i,j}$ by:
\[
  a_{i,j}:=  q^{i+j-1} \left (\qbinom{i+j-2}{i-1} + q \qbinom{i+j-1}{i} \right )+
  (1+q^i)\delta_{i,j}-\delta_{i,j+1}.
\]
Okada's crucial insight is that Theorem~\ref{qTSPPthm} holds if
\[
  \det(a_{i,j})_{1 \leq i,j \leq n}=
  \prod_{ 1 \leq i \leq j \leq k \leq n} {\left ( \frac{ 1-q^{i+j+k-1}}{1-q^{i+j+k-2}}\right )^2}=:b_n
  \qquad(n\geq 1).
\]
So for proving the $q$-TSPP conjecture, it is sufficient to prove this
conjectured determinant evaluation. For this purpose we apply a
computational approach originally proposed in~\cite{Z3} which is 
applicable to identities of the form
\[
  \det (a_{i,j})_{1 \leq i,j \leq n} = b_n\qquad(n\geq 1)
\] 
where $a_{i,j}$ and $b_n$ (with $b_n\neq0$ for all $n\geq 1$) are given
explicitly (as it is the case here).

The approach rests on the following induction argument on~$n$.  For
$n=1$, the identity is trivial.  Suppose the identity holds for $n-1$;
then the linear system
\[
  \begin{pmatrix}
    a_{1,1} & \cdots & a_{1,n-1} & a_{1,n} \\
    \vdots & \ddots & \vdots & \vdots \\
    a_{n-1,1} & \cdots & a_{n-1,n-1} & a_{n-1,n} \\
    0 & \cdots & 0 & 1
  \end{pmatrix}
  \begin{pmatrix}
    c_{n,1} \\
    \vdots \\
    c_{n,n-1} \\
    c_{n,n}
  \end{pmatrix}
  = 
  \begin{pmatrix}
    0 \\
    \vdots\\
    0\\
    1
  \end{pmatrix}
\]
has a unique solution $(c_{n,1},\dots,c_{n,n})$.
The component $c_{n,j}$ of this solution 
is precisely the $(n,j)$-cofactor divided by the $(n,n)$-cofactor 
of the $n\times n$-determinant. The division is meaningful because
the $(n,n)$-cofactor is just the $(n-1)\times(n-1)$-determinant,
which by induction hypothesis is equal to~$b_{n-1}$, which by general
assumption is nonzero. 
Since the $n\times n$-determinant can be expressed in terms of
the matrix entries $a_{n,j}$ and the normalized cofactors $c_{n,j}$
via
\[
  b_{n-1}\sum_{j=1}^n c_{n,j} a_{n,j},
\]
the induction step is completed by showing that this sum evaluates to~$b_n$.

The difficulty is that this last summation involves the
function~$c_{n,j}$ (of the discrete variables $n$ and $j$) for which we do not have an explicit expression for
general~$n$ and~$j$.  In order to achieve our goal, we guess a
suitable description of a function $c_{n,j}$ and then prove that it
satisfies the three identities
\begin{alignat}3
  c_{n,n}&=1&\qquad&(n\geq1),\label{eq:Normalization}\\
  \sum_{j=1}^n c_{n,j}a_{i,j}&=0&&(1\leq i<n),\label{eq:Soichi}\\
  \sum_{j=1}^n c_{n,j}a_{n,j}&=\frac{b_n}{b_{n-1}}&&(n\geq1)\label{eq:Okada}.
\end{alignat}
Once this is done, then by the argument given before the~$c_{n,j}$ must be 
precisely the normalized $(n,j)$-cofactors of the $n\times n$-determinant and
the determinant evaluation follows as a consequence. 
So in a sense, the function~$c_{n,j}$ plays the r\^{o}le of a
certificate for the determinant identity.


\section{The Odyssey}\label{secGuess}

In our the setting the certificate function $c_{n,j}$ will be described
implicitly by a system of linear recurrence equations in $n$ and $j$ with
coefficients depending polynomially on $q$, $q^j$ and~$q^n$.  Such recurrence
equations can be phrased as the elements of some noncommutative operator
algebras $\set Q(q,q^n,q^j)[S_n,S_j]$ where the symbols $S_x$ represent the
shifts $x\mapsto x+1$.  If a function is annihilated by certain operators (viz.\
it satisfies certain recurrence equations), then it is also annihilated by all
the elements in the (left) ideal generated by those operators.  We speak of an
annihilating ideal and represent such ideals by (left) Gr\"obner bases, so
that for instance ideal membership can be decided effectively.

The annihilating ideals we use for representing functions are such that they 
uniquely determine the function up to some finitely many initial values. 
Technically, this means that the ideals have dimension zero and that some 
particular polynomial coefficients appearing in the recurrence system must not vanish 
simultaneously for infinitely many points~$(n,j)$. 
These recurrence systems are similar but somewhat simpler than
$q$-holonomic systems~\cite{Z1}, which satisfy some additional requirements that 
are not needed for our proof.  We
have checked that all the ideals arising in our proof are indeed of the
desired form, but in the interest of clarity we suppress a more detailed
description of these checks here. For a complete analysis including all technical details we
refer to the supplementary material on our website.  Also in the interest of
clarity, we will from now on identify recurrence equations with their
corresponding operators.

A priori, there is no reason why the normalized cofactors $c_{n,j}$ should admit
a recursive description of the kind we are aiming at, but there is also no reason why they
should not. It turns out that they do, and this is fortunate because
for functions described in this way, techniques are known by which the
required identities~\eqref{eq:Normalization}, \eqref{eq:Soichi} and
\eqref{eq:Okada} can be proven algorithmically~\cite{Z1,T,CS,C}.
In order to find a recursive description for~$c_{n,j}$, we first
computed explicitly the normalized cofactors $c_{n,j}\in\set Q(q)$ for a few hundred
specific indices $n$ and $j$ by directly solving the linear system
quoted in Section~\ref{secReduction}.  Using an algorithm reminiscent
of polynomial interpolation, we then constructed a set of recurrences
compatible with the values of $c_{n,j}$ at the (finitely many) indices
we computed.

Polynomial interpolation applied to a finite sample $u_1,\dots,u_k$ of
an infinite sequence $u_n$ will always deliver some polynomial~$p$ of
degree at most $k-1$ which matches the given data. If it turns out
that this polynomial matches some further sequence terms
$u_{k+1},u_{k+2},\dots$, then it is tempting to conjecture that
$u_n=p(n)$ for all~$n$.  The more specific points~$n$ are found to
match, the higher is the evidence in favor of this conjecture, even
though without further knowledge no finite amount of points will ever
constitute a rigorous proof.

Very much analogously, it is possible to extract recurrence equations
from some finite number of values~$c_{n,j}$. The equations become
trustworthy if they also hold for points~$(n,j)$ which were not used for
their construction.  In this way, we have discovered a system of
potential recurrence equations for the~$c_{n,j}$, which, despite being
respectable in size (about 30 Megabytes), appears to be a rather
plausible candidate for a recursive description of the normalized
cofactors~$c_{n,j}$. The system is available for download on our
website.

\section{The Nostos}\label{secProof}

Now we switch our point of view. We discard the definition that
$c_{n,j}$ be the normalized $(n,j)$-cofactor and redefine the discrete
function $c_{n,j}$ as the unique solution of the guessed recurrence
system whose (finitely many) initial values agree with the normalized
$(n,j)$-cofactor. If we succeed in proving that $c_{n,j}$ defined in
this way satisfies the identities~\eqref{eq:Normalization},
\eqref{eq:Soichi} and \eqref{eq:Okada}, then we are done.
For this purpose we
provide operators which belong to the annihilating
ideal of $c_{n,j}$ or related ideals and have certain features which imply
the desired identities. So in a sense, these operators play the r\^ole
of certificates for the identities under consideration. 

Because of their astronomical size (up to 7~Gigabytes; equivalent to
more than one million printed pages; corresponding to about 2.5~tons
of paper), these certificates are not included explicitly in this
article but provided only electronically on our website. Also because
of their size, it was not possible to construct them by
simply applying the standard algorithms from~\cite{Z1,T,C,CS}. A
detailed explanation of how exactly we found the certificates is
beyond the scope of this article and will be given in a separate
publication.  But this lack of explanation does not at all affect the
soundness of our proof,
because the correctness of the certificates can be checked
independently by simply performing ideal membership tests.  
Our certificates are so big that even this ``simple'' calculation is
not quite trivial, but a reader with a sufficient amount of patience
and programming expertise will be able to do it.

We proceed by explaining the properties of the certificates provided
on our website and why they imply \eqref{eq:Normalization},
\eqref{eq:Soichi} and \eqref{eq:Okada}. 


\subsection{A certificate for \eqref{eq:Normalization}} 

To certify that $c_{n,n}=1$ for all $n\geq 1$, we provide a recurrence in
the annihilating ideal of $c_{n,j}$ which is of the special form
\[
  p_7(q,q^j,q^n)c_{n+7,j+7}+\dots+p_1(q,q^j,q^n)c_{n+1,j+1}+p_0(q,q^j,q^n)=0.
\]
By virtue of the substitution $j\mapsto n$, it translates into a
recurrence for the diagonal sequence~$c_{n,n}$.  This recurrence
contains the annihilating operator $S_n-1$ of the constant
sequence~$1$ as a (right) factor, and therefore, after checking
$c_{1,1}=c_{2,2}=\cdots=c_{7,7}=1$, it can be concluded that
$c_{n,n}=1$ for all~$n$.

Similarly, we showed that $c_{n,0}=0$ for all $n\geq 1$ and that
$c_{n,j}=0$ for all $j>n$. This knowledge greatly simplifies the
following proofs of the summation identities.

\subsection{Certificates for \eqref{eq:Soichi}} 

In order to prove the first summation identity, we
translate~\eqref{eq:Soichi} into the equivalent formulation
\begin{equation}\label{Soichi1}
  \sum_{j=1}^n \frac{q^{i+j-1}(q^{i+j}+q^i-q-1)}{q^i-1}\qbinom{i+j-2}{i-1}c_{n,j} =
  c_{n,i-1}-(q^i+1)c_{n,i}\quad(1\leq i<n),
  \tag{$\ref{eq:Soichi}'$}
\end{equation}
taking into account that $c_{n,0}=0$.  We provide certificates
for~\eqref{Soichi1}.

To this end, let~$c'_{n,i,j}$ be the summand of the sum on the
left-hand side.  A recursive description for $c'_{n,i,j}$ can be
computed directly from the defining equations of $c_{n,j}$ and the
fact that the rest of the summand is a $q$-hypergeometric factor.  In
the corresponding operator ideal~$I'$ we were able to find two
different recurrence equations for $c'_{n,i,j}$ which are of the special
form
\[
  p_{4,3}c'_{n+4,i+3,j}+\dots +p_{1,0}c'_{n+1,i,j}+p_{0,1}c'_{n,i+1,j}+p_{0,0}c'_{n,i,j} =
  t_{n,i,j+1} - t_{n,i,j}
\]
where the $p_{4,3},\dots,p_{0,0}$ are rational functions in
$\set{Q}(q,q^i,q^n)$ and $t_{n,i,j}$ is a
$\set{Q}(q,q^i,q^j,q^n)$-linear combination of certain shifts of
$c'_{n,i,j}$ (which are determined by the Gr\"obner basis of~$I'$
as described in~\cite{C}). Next observe that $c'_{n,i,j}=0$ for $j\leq 0$ (because
of the $q$-binomial coefficient) and also for $j>n$ (because
$c_{n,j}=0$ for $j>n$). When summing the two recurrence equations for $j$ from $-\infty$
to $+\infty$, the right-hand side telescopes to~$0$ and the left-hand
side turns into a recurrence for the sum in~\eqref{Soichi1}. This
method is known as creative telescoping.  Finally we obtain two
annihilating operators~$P_1$ and $P_2$ for the left-hand side
of~\eqref{Soichi1}.

For the right-hand side of \eqref{Soichi1}, we can again construct an
ideal of recurrences from the defining equations of~$c_{n,j}$.  It
turns out that this ideal contains $P_1$ and~$P_2$, so that both sides
of \eqref{Soichi1} are annihilated by these two
operators. Additionally, $P_1$ and $P_2$ have been constructed such
that they require only finitely many initial values to produce a
uniquely determined bivariate sequence.  Therefore it suffices to
check that the two sides of \eqref{Soichi1} agree at some finitely
many points $(n,j)$.

\subsection{The certificate for \eqref{eq:Okada}} 

To certify the final identity, we rewrite \eqref{eq:Okada} equivalently into
\begin{equation}\label{Okada1}
\tag{$\ref{eq:Okada}'$}
 (1+q^n)-c_{n,n-1}+\sum_{j=1}^n \frac{q^{n+j-1}(q^{n+j}+q^n-q-1)}{q^n-1}\qbinom{n+j-2}{n-1}c_{n,j}
 =\frac{(q^{2n};q)_n^2}{(q^n;q^2)_n^2}
 \qquad(n\geq 1)
\end{equation}
where $(a;q)_n:=(1-a)(1-qa)\cdots(1-q^{n-1}a)$ denotes the
$q$-Pochhammer symbol.  As before, we use creative telescoping to
provide a certified operator~$P$ which annihilates the sum on the
left-hand side. In the present case, a single operator is sufficient
because the sum depends only on a single variable~$n$ (there is no~$i$
there).  Using the operator~$P$ and the defining equations for~$c_{n,j}$, 
we then construct a recurrence for the entire left-hand
side, which turns out to have order twelve.  As this recurrence is a left multiple
of the second order operator annihilating the right-hand side, the
proof of Theorem~\ref{qTSPPthm} is completed by
checking~\eqref{Okada1} for $n=1,2,\dots,12$.

\section{Epilog}\label{Conclusion}

Paul Erd{\H o}s famously (and naively) believed that every short and elegant mathematical statement
has a short and elegant ``proof from the book'', and if humans tried hard enough, they
would eventually find it. Kurt G\"odel, on the other hand, meta-proved that there exist
many short statements whose shortest possible proof is very long. It is very possible that
the q-TSPP theorem {\it does} have a yet-to-be-found {\it proof from the book}, but it
is just as possible that it does not, and while we are sure that the present proof is not
the shortest-possible (there were lots of random choices in designing the proof), it may well
be the case that the shortest-possible proof is still very long, and would still require heavy-duty
computer calculations.

Be that as it may, we believe that our general {\it approach}, and our way of {\it taming} the
computer to prove something that seemed intractable with today's hardware and software are
very elegant, and deserve to be included in the {\it book}.

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\end{document}
